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AC power describes energy transfer associated with time-varying electric potential difference and electric current, including the distinction between instantaneous transfer and its average. With the passive sign convention, current enters the terminal marked positive, and instantaneous power absorbed by the element is $p(t)=v(t)i(t)$. A positive value means energy entering the element; a negative value means energy returning through that terminal pair.

For sinusoidal steady operation, choosewhere $V$ and $I$ are root mean square magnitudes. The cosine-product identity givesAveraging over a complete period removes the second term, giving active power $P=VI\cos \phi$. This derivation separates the energy-transfer average from a twice-frequency oscillation.^[ref]

For an original example, $V=120\mathrm{V}$, $I=5\mathrm{A}$ and $\cos \phi =0.8$ give active power480W. The oscillating term has amplitude600W, so instantaneous power ranges from−120 to1080W. Over ten complete cycles at50Hz, the net energy absorbed is $480(10/50)=96\mathrm{J}$. Temporary negative power can accompany positive net transfer when an element stores and returns energy.

The product $VI=600$ in volt-amperes is the apparent power. In the single-frequency sinusoidal convention, reactive power is $Q=VI\sin \phi$, giving360 in reactive volt-amperes for a lagging current with $\sin \phi =0.6$. Complex electrical power packages $P$ and $Q$, while the power factor $P/(VI)=0.8$ distinguishes net transfer from the terminal magnitude product. The primary MIT treatment derives these quantities and makes the peak-versus-RMS convention explicit.^[ref]

A measurement must retain the relative phase of the two channels. Multiplying separately measured RMS magnitudes gives apparent power, not generally active power. For nonsinusoidal signals, the time average of $v(t)i(t)$ remains the definition of active power, but one phase angle need not describe every harmonic contribution.

Active power, also called real power, is the time average of instantaneous power transferred through a terminal pair. For periodic electric potential difference $v(t)$ and electric current $i(t)$ of common period $T$, the passive sign convention givesThe net energy entering during each period is $PT$. Positive active power denotes net absorption under this convention; it can represent useful conversion to other forms of energy as well as heating.

For sinusoids of RMS magnitudes $V$ and $I$ with relative phase $\phi$, $P=VI\cos \phi$. For a pure resistor with constant resistance $R$, the stronger waveform-independent statement follows from $v(t)=Ri(t)$:This is why the root mean square current is the constant-current value giving the same heating in that resistor.^[ref]

An original harmonic example shows why RMS values alone do not determine active power. Letin volts and amperes. Orthogonality over a period gives $I_{\mathrm{rms}}=\sqrt{3^2+4^2}=5\mathrm{A}$, but the third harmonic averages to zero when multiplied by the fundamental potential difference. Consequently $P=100(3)=300\mathrm{W}$, whereas apparent power is $100(5)=500\mathrm{VA}$. The power factor is0.6 even though the fundamental current is exactly in phase with the fundamental potential difference.

More generally, Fourier analysis of real periodic waveforms with RMS harmonic magnitudes giveswhen the expansion and average are justified. Cross terms of different harmonic orders vanish. A single fundamental-frequency complex electrical power therefore need not capture every harmonic contribution to the total average.

For an ideal inductor or capacitor in periodic steady operation, the stored energy returns to its starting value each cycle, so its net active power is zero despite nonzero exchange during the cycle. Real components can also have losses. The primary MIT treatment explicitly separates resistor dissipation from periodic inductive and capacitive energy exchange.^[ref]

Complex electrical power represents the active and reactive parts of sinusoidal electrical energy transfer in one complex number. Using phasors $V$ and $I$ whose magnitudes are root mean square values, with the passive sign convention, defineThe star denotes a complex conjugate. The real part is active power; the imaginary part is reactive power. The magnitude $|S|=|V||I|=\sqrt{P^2+Q^2}$ is apparent power. A source using peak-amplitude phasors instead has the factor $1/2$: $S=V_{\mathrm{peak}}{I}_{\mathrm{peak}}^{\ast }/2$.^[ref]

To see why the conjugate matters, write $V=|V|e^{j\alpha }$ and $I=|I|e^{j\beta }$. Then $S=|V||I|e^{j(\alpha -\beta )}$ depends on the relative phase. Changing the common time reference multiplies both phasors by $e^{j\theta }$, leaving $V{I}^{\ast }$ unchanged. The product $VI$ would instead acquire $e^{2j\theta }$ and would not represent a time-reference-independent average-power quantity.

For an original example, choose $V=100\mathrm{V}$ and $I=(4-j3)\mathrm{A}$. ThenThus active power is400W, reactive power is300 reactive volt-amperes and power factor is0.8. The corresponding electrical impedance is $V/I=16+j12\Omega$. With RMS current magnitude5A, the resistive term absorbs $16(5^2)=400\mathrm{W}$, agreeing with the real part.

Under the stated sign convention and $e^{j\omega t}$ time convention, an inductive reactance gives positive $Q$ and a capacitive reactance negative $Q$. Negative $Q$ does not mean a negative dissipated-energy average. For example, adding an ideal parallel capacitive branch contributing $-j300\mathrm{VA}$ cancels this example's reactive part, leaving terminal complex power400VA and RMS source current4A at100V, while the original resistive dissipation stays400W.

This one-frequency representation must not be applied indiscriminately to a distorted waveform. Different harmonics need their own phase relationships; the definition of active power remains the time average of the actual instantaneous power.

Electrical impedance is the ratio of terminal electric potential difference and electric current phasors at one specified positive angular frequency $\omega$. For an electrical circuit modeled as a linear time-invariant system, use the sinusoidal steady state after any decaying transient has settled. With the passive sign convention, positive electric current enters the terminal assigned positive electric potential difference. Define peak complex amplitudes byThe physical $v(t)$ and $i(t)$ are real; the complex numbers encode their amplitudes and wave phases. The unit of electrical impedance is the ohm, $1\Omega =1\mathrm{V}/\mathrm{A}$. Both phasors may instead use root mean square magnitudes, with the same ratio $Z$. A nonlinear diode can generate additional frequencies; a single $Z(\omega )$ then does not describe the whole response.

Write the electrical impedance in Cartesian and polar forms asHere $R$ is the resistive real part, $X$ the signed electrical reactance, and $\varphi$ the voltage-minus-current phase difference, defined modulo a full turn. The Pythagorean theorem gives the modulus from a right triangle in the complex plane. The complex argument gives the rotation from the current phasor to the voltage phasor. Thus $V_{\mathrm{rms}}=|Z|I_{\mathrm{rms}}$, while negative $\varphi$ means electric current leads terminal electric potential difference. For $R>0$, $\varphi =\arctan (X/R)$ lies between $-\pi /2$ and $\pi /2$; outside that case, use the quadrant of the complex number rather than a bare inverse tangent. The geometric magnitude and phase difference are the two pieces of the AC relation.^[ref]

Multiplication by $Z$ acts on phasors, not on instantaneous physical signals. For example, when $i(t)=I_0\cos \omega t$,The second term is in phase quadrature with electric current; it cannot be represented by pointwise multiplication $v(t)=Zi(t)$ when $X\ne 0$. This is the useful compression: one complex multiplication replaces separate amplitude and phase difference calculations, while the real part restores the physical signal.

An ideal resistor, inductor and capacitor have $Z_R=R$, $Z_L=\mathrm{i}\omega L$ and $Z_C=1/(\mathrm{i}\omega C)$. Their electrical impedances add in a series RLC circuit, giving $Z=R+\mathrm{i}(\omega L-1/(\omega C))$. The current amplitude is $V_0/|Z|$, so cancellation of the imaginary parts explains the resonant series RLC response. The component-law derivation and opposite complex exponential convention are in reactance phase convention.

The harmonic active power sum expresses average electrical power as the sum of contributions from matching components of periodic terminal electric potential difference and electric current. With a common fundamental frequency and root mean square magnitudes for nonzero harmonic components, writeHere $V_0,I_0$ are the signed DC components. The passive sign convention makes the average product absorbed active power. Averaging over a complete fundamental period givesFor finite sums this follows from the cosine-product identity: products of different integer harmonic components average to zero, and matching components give half the cosine of their relative phase. For square-integrable periodic waveforms, the corresponding Fourier series identity follows by continuity of the inner product in the mean-square norm. The orthogonality is a property of the averaging interval and basis, not an assumption that the physical sources are independent.

For an original example, set $V_0=10\mathrm{V}$, $I_0=2\mathrm{A}$, $V_1=100\mathrm{V}$, $I_1=3\mathrm{A}$, and fundamental relative phase $\pi /3$. Add third-harmonic magnitudes $V_3=20\mathrm{V}$, $I_3=1\mathrm{A}$ with relative phase $\pi$. ThenThe third-harmonic contribution is $-20\mathrm{W}$, even though the total absorbed active power is positive. The root mean square values satisfy $V_{\mathrm{rms}}^2=10500{\mathrm{V}}^2$ and $I_{\mathrm{rms}}^2=14{\mathrm{A}}^2$, so apparent power is about $383.41\mathrm{VA}$ and total power factor about $0.3912$.

For a purely sinusoidal supply with no DC component, only $V_1$ survives, giving $P=V_1{I}_1\cos ({\alpha }_1-{\varphi }_1)$ even if the electric current is distorted. Current in other harmonic components can still increase conductor heating. The usual one-frequency complex electrical power construction is derived in the primary MIT treatment; the sum above applies that average-product reasoning separately to orthogonal components.^[ref]

A proposed measurement records the two waveforms synchronously and compares their directly averaged product with the resolved sum. A non-integer-period window, unresolved harmonic components, channel timing error or changing waveform can spoil the simple finite-record equality. The formula assumes a shared periodic model; multiplying separately measured RMS magnitudes gives apparent power instead.

James L. Kirtley Jr. is an electrical engineer whose work concerns electric machines and electric power systems. MIT's departmental profile identifies him as a professor emeritus of electrical engineering, with energy and robotics among his research areas.^[ref]

The engineering problem is to match a motor's behaviour to its application. MIT reported in 2016 that Kirtley worked on customized electric motors for the MIT Cheetah robot, where brief peaks of torque serve running and jumping. The same interview describes his interest in electric motors interacting with microgrids, particularly their dynamic behaviour during supply disturbances.^[ref] A short burst and sustained operation place different demands on the system; the largest instantaneous output alone does not describe its operating role.

His AC Power Flow in Linear Networks by James L. Kirtley Jr. connects the phasor representation of electric potential difference and electric current to physical energy transfer. In its peak-amplitude convention, complex electrical power is $S=V_{\mathrm{pk}}{I}_{\mathrm{pk}}^{\ast }/2$. With RMS phasors $V=V_{\mathrm{pk}}/\sqrt{2}$ and $I=I_{\mathrm{pk}}/\sqrt{2}$, this becomes $S=V{I}^{\ast }$. The complex conjugate ensures that the relative phase enters the active power.^[ref]

For an original load example in sinusoidal steady state, take $V=20\mathrm{V}$ and electrical impedance $Z=(3+4i)\Omega$, using RMS phasors and the passive sign convention: electric current enters the positive-potential terminal. A series resistor and inductor realize this model at the specified angular frequency when $R=3\Omega$ and $\omega L=4\Omega$. ThenThus active power is 48 watts, reactive power is 64 vars, apparent power is 80 volt-amperes, and power factor is $48/80=0.6$. Write $\varphi =\arctan (4/3)$ and $x=\omega t$. The corresponding electric potential difference and electric current areMultiplying and applying a trigonometric identity gives instantaneous powerIts cycle average is 48 watts, while its range is $[-32,128]$ watts. The negative portions represent temporary return of stored energy from this ideal load; the electrical resistance still dissipates a positive average. A synchronized product of the two waveforms reveals this behaviour, whereas multiplying their root mean square magnitudes gives only the apparent power.

Joule heating is the dissipative conversion of electrical work into internal energy in an electrical conductor. In a metal, the electric field supplies energy to charge carriers, which transfer it to the material. This makes an immersed resistor a useful calibrated heater and makes electrical resistance a source of losses in wiring. Wind-driven electric currents are a proposed source of heating of an inflated giant planet.

For a stationary, purely resistive Ohmic conductor, use the passive sign convention: $V$ is the terminal electric potential difference from the entry to the exit of positive electric current $I$. The absorbed electrical power and Ohm's law giveThe last two expressions assume $V=IR$ at the relevant material temperature. For constant electrical resistance and electric current over a duration $\Delta t$, the dissipated energy is $E_{\mathrm{diss}}=I^{2}R\Delta t$; otherwise use $E_{\mathrm{diss}}=\int V(t)I(t)dt$. In SI, $\mathrm{V}\mathrm{A}=\mathrm{W}=J/s$.^[ref]

Absorbed electrical power $VI$ is more general than Joule heating: a capacitor can store energy, a battery can change its chemical energy, and an electrical circuit can deliver mechanical work or emit light. Even a heated resistor need not show an indefinitely rising temperature: in a thermal steady state, its dissipated power leaves by heat transfer. With the heater inside a calorimeter boundary, electrical input crosses as work; the subsequent heat transfer to the room is a different term in the first law of thermodynamics.

A lagging power factor describes an absorbing sinusoidal electrical load whose electric current lags its terminal electric potential difference. With the passive sign convention, chooseThe current maximum occurs later by $\phi /\omega$. The RMS phasors give complex electrical powerIn this equation the supply phasor is the real number $V$, while $I^{\ast }$ denotes the conjugate of the current phasor. Positive $Q$ is inductive reactive power under this convention. The primary MIT notes derive the conjugation and the peak-versus-RMS factor.^[ref]

For an original $230\mathrm{V}$, $10\mathrm{A}$ example with $\cos \phi =0.8$, active power is $1840\mathrm{W}$ and reactive power is $1380\mathrm{var}$. At $50\mathrm{Hz}$, $\phi \approx 0.6435$ radians corresponds to a current delay of about $2.05\mathrm{ms}$. The factor $0.8$ gives the magnitude of the phase relationship, while the word lagging supplies the direction lost by the cosine.

An ideal inductor is the limiting case $\phi =\pi /2$: it alternately stores and returns magnetic energy, with zero mean active power. An absorbing resistive-inductive load usually has both positive active power and positive reactive power. An ideal parallel capacitor can reduce the source's quadrature electric current; sufficient extra capacitance eventually crosses into leading power factor rather than improving the factor indefinitely.

For distorted waveforms, specify whether lagging refers to the fundamental displacement power factor; one phase angle does not characterize arbitrary harmonic components. A signed meter display is also not a universal lead/lag label. The manufacturer's sign-convention explanation distinguishes phase direction from the convention used for displayed factor signs.^[ref]

A leading power factor describes an absorbing sinusoidal electrical load whose electric current leads its terminal electric potential difference. Under the passive sign convention, writeThe current maximum occurs earlier by $\theta /\omega$. The complex electrical power isThe negative reactive power is consistent with a capacitive terminal response. It does not imply negative mean absorbed active power.

For an original $100\mathrm{V}$, $2\mathrm{A}$ example with $\theta =\pi /3$, the power factor magnitude is $0.5$, active power is $100\mathrm{W}$, and reactive power is $-100\sqrt{3}\approx -173.21\mathrm{var}$. At $50\mathrm{Hz}$ the current leads by $1/(6f)\approx 3.33\mathrm{ms}$. A load with the same RMS magnitudes and the opposite phase displacement has identical active power but a lagging power factor and opposite reactive power.

For an ideal capacitor, $i=Cdv/dt$ gives a quarter-cycle current lead. Its mean active power is zero, although instantaneous power alternates as electrical energy enters and leaves the field. The primary one-frequency power derivation separates this quadrature exchange from net energy transfer.^[ref]

An inductive load can acquire a leading power factor at the source when power factor correction supplies more capacitive reactive power than the load absorbs. The current reduction interval for shunt reactive compensation shows why mildly leading operation can still draw less source electric current than no correction, while excessive correction reverses that benefit.

Record phase direction or signed reactive power together with the numerical factor. The same positive cosine can describe leading and lagging loads, and a meter may assign a separate sign according to its display convention.^[ref] For nonsinusoidal signals, state whether leading describes the fundamental displacement power factor; the complete power factor also depends on waveform distortion.

The passive sign convention defines terminal electric potential difference $V=\varphi (A)-\varphi (B)$ and positive electric current entering at $A$. Then $VI$ is absorbed electrical power: positive means the component receives energy, negative means it supplies energy. Reversing both reference directions leaves the product unchanged. For a positive electrical resistance obeying Ohm's law, both $V$ and $I$ reverse together and Joule heating remains $I^{2}R\ge 0$.

For the imposed electric current $I=I_0\cos \omega t$, Faraday's law gives the self-induced electromotive forceThe externally applied terminal voltage, with the passive sign convention, has the opposite sign:Thus the terminal voltage leads the electric current by $\pi /2$. The instantaneous input power isAn ideal inductor alternately stores magnetic energy and returns it to the source. Real winding electrical resistance would add mean Joule heating $I_0^2{R}_w/2$; the ideal zero result assumes that loss is neglected.

Faraday's law of induction relates an induced electromotive force to the rate of change of magnetic flux. Choose a positive orientation around a loop and its associated surface normal by the right-hand rule. For a stationary conducting loop,Unlike an electrostatic field, the induced electric field generally has nonzero curl; the differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a moving circuit, magnetic flux transport for a moving circuit includes the magnetic Lorentz force on the charges:where $\mathbf{u}$ is the local wire velocity. Thus changing the field or moving a conductor can produce induction.

The minus sign is Lenz's law: an induced electric current, when a closed conducting path permits it, produces a magnetic field opposing the change of linked flux. It need not oppose the original field itself. For example, increasing flux through a loop produces a current tending to reduce that increase; decreasing the same flux reverses the current. The opposition is consistent with conservation of energy: external mechanical or electrical work is needed to maintain a change against its inductive reaction.

For a coil of $N$ turns, define the magnetic flux linkage $\Lambda ={\sum }_{j=1}^N{\Phi }_j$, reducing to $N{\Phi }_B$ when every turn links the same flux. For fixed geometry and a linear magnetic response, self-inductance $L$ is defined by $\Lambda =LI$, and the coil's induced electromotive force isThe corresponding terminal voltage drop, with the passive sign convention, is $V_L=LdI/dt$. It opposes changes in electric current; it does not require the current itself to vanish.

Building the current from zero requires power $P=I{V}_L=LIdI/dt$. Integrating gives the energy stored by an ideal inductor:This energy is stored in the magnetic field and can be returned when the current falls. In a series RL circuit driven by a constant voltage $V_0$, Kirchhoff's voltage law gives $L\dot{I}+RI=V_0$. Starting with zero current, $I=(V_0/R)(1-e^{-tR/L})$, with time constant $L/R$. This makes the resistance to rapid current changes directly observable. Two coils also have mutual inductance: changing primary current induces secondary electromotive force $-Md{I}_1/dt$ with a polarity determined by the chosen winding orientations, the principle behind a transformer.

Reactive power is the imaginary part of complex electrical power in sinusoidal steady operation. With RMS phasors, the passive sign convention and $e^{j\omega t}$ time convention,where $\phi$ is potential-difference phase minus current phase. Inductive current lag gives positive $Q$, while capacitive current lead gives negative $Q$. Its conventional unit is the reactive volt-ampere, written var. These definitions retain the specified phasor and sign conventions.^[ref]

The physical role is periodic energy exchange rather than an additional dissipated-energy average. For an ideal inductor, $v=Ldi/dt$ and stored energy is $w=L{i}^2/2$, so $p=dw/dt$. Its energy returns to its initial value after a steady cycle and mean absorbed power is zero. Nevertheless,For an ideal capacitor, $Q_C=-\omega C{V}_{\mathrm{rms}}^2$, with the negative sign representing the opposite exchange phase. Primary circuit derivations explicitly separate these exchanges from resistor dissipation.^[ref]

For an original example, $L=20\mathrm{mH}$, $I_{\mathrm{rms}}=5\mathrm{A}$ and $f=50\mathrm{Hz}$ give $Q_L\approx 157.08\mathrm{var}$. Peak stored energy is0.50J and mean stored energy0.25J, consistent with $2\omega ⟨w⟩$. Zero ideal average dissipation therefore does not mean zero current or zero energy storage.

For a separate ideal single-frequency compensation example, a100V load with $S=400+j300\mathrm{VA}$ can have its source reactive component canceled by a parallel capacitor satisfyingThe capacitor contributes $-j300\mathrm{VA}$, leaving400W of source active power and4A RMS source current. This illustrates power factor correction under the stated ideal assumptions, rather than selecting a real component or proving its ratings.

Reactive exchange can increase the current carried through resistive wiring and thereby affect real losses. For distorted waveforms, definitions of non-active power need additional care; the one-frequency imaginary-part definition should not silently be extended to every waveform.

Reactive power overcompensation occurs when capacitive correction exceeds the inductive reactive power that it is intended to cancel. Under the passive sign convention and a positive-frequency sinusoidal supply, the source then sees negative net reactive power, producing a leading power factor. This can happen when a fixed capacitor bank remains connected as the inductive load falls. Eaton's application paper discusses excessive leading operation and the use of switched correction stages.^[ref]

Consider an ideal fixed RMS electric potential difference $V$ at angular frequency $\omega$, supplying a load that absorbs constant $P>0$ and $Q_L>0$. Let $Q_{\mathrm{cap}}=\omega C{V}^2\ge 0$ be the magnitude of the correction's reactive power; its signed contribution is $-Q_{\mathrm{cap}}$. Adding ideal shunt capacitance givesThe model holds the load and supply fixed while changing $C$. It omits losses and additional network dynamics. Optimal shunt capacitance for sinusoidal power factor correction follows by minimizing the squared root mean square electric current:For $C<C_{\ast }$ the remaining reactive power is positive and the power factor is lagging. For $C>C_{\ast }$ it is negative and the power factor is leading. The overcompensation threshold is $C_{\ast }$, but the current reduction interval for shunt reactive compensation is wider. Comparing the square of the corrected and uncorrected electric current givesThus source RMS current is reduced for $0<C<2C_{\ast }$, equals the uncorrected value at $C=2C_{\ast }$, and is larger for $C>2C_{\ast }$. An overcompensated operating point between $C_{\ast }$ and $2C_{\ast }$ still draws less source electric current than the uncorrected load.

For an original example, keep the parent article's chosen $P=1000\mathrm{W}$, $Q_L=750\mathrm{var}$, $V=200\mathrm{V}$ and 50 hertz. The uncorrected source draws 6.25 amperes at power factor0.8. Full cancellation needs $C_{\ast }\approx 59.68\mu \mathrm{F}$ and reduces source electric current to 5 amperes. A 100 microfarad correction gives $Q_{\mathrm{net}}\approx -506.64\mathrm{var}$ and 5.605 amperes, with a leading factor about0.8920. A 150 microfarad correction instead gives about $-1134.96\mathrm{var}$ and7.563 amperes, worse than no correction. The crossing is $2C_{\ast }\approx 119.37\mu \mathrm{F}$.

The optimum changes with the load. If the example load's $P$ and $Q_L$ both halve while its original optimal correction remains connected, then $P=500\mathrm{W}$ and $Q_{\mathrm{net}}=-375\mathrm{var}$. Source electric current is 3.125 amperes and the factor is 0.8 leading. This dependence motivates automatic power factor correction. The cited controller description measures system electric potential difference and electric current, using external current transformers, and switches discrete capacitor bank stages to follow a target.^[ref]

Retain the phase or sign of $Q_{\mathrm{net}}$ when reporting the result. A bare power factor magnitude does not distinguish leading from lagging; the manufacturer FAQ also illustrates how a meter's signed display depends on its convention.^[ref] The calculation assumes sinusoidal steady state; current distortion factor is a separate consideration for nonsinusoidal current.

With the passive sign convention applied to the network, its entering electric current is $J=-I$, and the same relation becomes $V=V_{\mathrm{oc}}+J{R}_{\mathrm{th}}$. The electrical load uses the passive sign convention with entering electric current $I$, so it absorbs power $VI$. A positive delivered electric current therefore produces a voltage drop for positive $R_{\mathrm{th}}$; changing the reference direction without changing the equation would give the wrong sign.

The reason is linearity of Ohm's law, Kirchhoff's current law, Kirchhoff's voltage law, and the constitutive relations of any linear dependent electrical sources. Solve once at $I=0$ with the independent electrical sources present. Subtract this solution from a solution with prescribed delivered electric current $I$: the fixed independent electrical sources cancel, leaving a homogeneous linear network whose terminal voltage change is proportional to $I$. This is a short superposition principle argument for an affine function; the fixed offset need not vanish.

The qualification about a well-defined response matters. An ideal voltage source alone has $R_{\mathrm{th}}=0$. An ideal nonzero current source alone constrains $I$ and does not have a finite open-circuit voltage or a representation $V=V_{\mathrm{oc}}-I{R}_{\mathrm{th}}$ with finite coefficients. Linear dependent electrical sources can give negative resistance, and singular ideal networks may fail to have a unique finite response. Neither a finite nor a strictly positive passive $R_{\mathrm{th}}$ follows merely from calling an electrical circuit linear. ^[ref]